Question

The net ionic equation for the Breathalyzer test used to indicate alcohol concentration in the body is as follows: \(16 \mathrm{H}^{+}(a q)+2 \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+3 \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}(a q) \longrightarrow\) $$3 \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}(a q)+4 \mathrm{Cr}^{3+}(a q)+11 \mathrm{H}_{2} \mathrm{O}(l)$$ (a) How many grams of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) must be used to consume \(1.50 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O} ?\) (b) How many grams of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\) can be produced from \(80.0 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O} ?\)

Short answer

(a) 6.38 g of \( \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 \); (b) 104.29 g of \( \mathrm{C}_2\mathrm{H}_4\mathrm{O}_2 \).

Step-by-step solution

Calculate Moles of C₂H₆O

We need to find out how many moles of \( \mathrm{C}_2\mathrm{H}_6\mathrm{O} \) are in 1.50 g. The molar mass of \( \mathrm{C}_2\mathrm{H}_6\mathrm{O} \) is calculated as: \( (2 \times 12.01) + (6 \times 1.01) + 16.00 = 46.08 \, \text{g/mol} \). So, the moles of \( \mathrm{C}_2\mathrm{H}_6\mathrm{O} \) are: \[ \text{Moles of } \mathrm{C}_2\mathrm{H}_6\mathrm{O} = \frac{1.50 \, \text{g}}{46.08 \, \text{g/mol}} = 0.0326 \, \text{mol} \]

Use Stoichiometry to Find Moles of K₂Cr₂O₇

The balanced equation shows that 2 moles of \( \mathrm{Cr}_2\mathrm{O}_7^{2-} \) react with 3 moles of \( \mathrm{C}_2\mathrm{H}_6\mathrm{O} \). Using stoichiometry: \[ \text{Moles of } \mathrm{Cr}_2\mathrm{O}_7^{2-} = \frac{2}{3} \times 0.0326 \, \text{mol} = 0.0217 \, \text{mol} \] For \( \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 \), which contains \( \mathrm{Cr}_2\mathrm{O}_7^{2-} \), this is also 0.0217 moles.

Calculate Grams of K₂Cr₂O₇

The molar mass of \( \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 \) is calculated as: \( (2 \times 39.10) + (2 \times 51.996) + (7 \times 16.00) = 294.18 \, \text{g/mol} \). So, the grams needed are: \[ \text{Grams of } \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 = 0.0217 \, \text{mol} \times 294.18 \, \text{g/mol} = 6.38 \, \text{g} \]

Calculate Moles of C₂H₆O for 80.0g

Now, we find out how many moles of \( \mathrm{C}_2\mathrm{H}_6\mathrm{O} \) are in 80.0 g: \[ \text{Moles of } \mathrm{C}_2\mathrm{H}_6\mathrm{O} = \frac{80.0 \, \text{g}}{46.08 \, \text{g/mol}} = 1.736 \, \text{mol} \]

Use Stoichiometry to Find Moles of C₂H₄O₂

According to the balanced equation, 3 moles of \( \mathrm{C}_2\mathrm{H}_6\mathrm{O} \) produce 3 moles of \( \mathrm{C}_2\mathrm{H}_4\mathrm{O}_2 \). Therefore, the moles of \( \mathrm{C}_2\mathrm{H}_4\mathrm{O}_2 \) are the same as moles of \( \mathrm{C}_2\mathrm{H}_6\mathrm{O} \): \( 1.736 \, \text{mol} \).

Calculate Grams of C₂H₄O₂ Produced

The molar mass of \( \mathrm{C}_2\mathrm{H}_4\mathrm{O}_2 \) is: \( (2 \times 12.01) + (4 \times 1.01) + (2 \times 16.00) = 60.05 \, \text{g/mol} \). So, the grams produced are: \[ \text{Grams of } \mathrm{C}_2\mathrm{H}_4\mathrm{O}_2 = 1.736 \, \text{mol} \times 60.05 \, \text{g/mol} = 104.29 \, \text{g} \]

Key concepts

Molar Mass Calculation

To solve problems involving chemical reactions, calculating the molar mass is a crucial first step. Molar mass is the mass of one mole of a compound, allowing for conversion between mass and moles.

• To find the molar mass of ethanol \( \mathrm{C}_2\mathrm{H}_6\mathrm{O} \), sum the atomic masses of its elements: \((2 \times 12.01) + (6 \times 1.01) + 16.00 = 46.08 \, \text{g/mol}\).
• Similarly, for potassium dichromate \( \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 \), use \((2 \times 39.10) + (2 \times 51.996) + (7 \times 16.00) = 294.18 \, \text{g/mol}\).
• This step links mass and moles, facilitating further calculations in stoichiometry.

Net Ionic Equation

In a chemical reaction, the net ionic equation illustrates only the components involved in the reaction, excluding spectator ions. It simplifies the reaction process to focus on actual changes.

• In the context of a Breathalyzer test, the net ionic equation is crucial for identifying how ethanol is oxidized to acetic acid, forming water and chromium ions.
• The given equation shows ethanol's interaction with dichromate ions and protons: \( 16 \mathrm{H}^{+} + 2 \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 3 \mathrm{C}_2\mathrm{H}_6\mathrm{O} \rightarrow 3 \mathrm{C}_2\mathrm{H}_4\mathrm{O}_2 + 4 \mathrm{Cr}^{3+} + 11 \mathrm{H}_2\mathrm{O} \).
• This highlights how ethanol is oxidized by chromium ions, which is essential for calculating stoichiometry.

Reaction Stoichiometry

Reaction stoichiometry allows us to predict the amount of reactants and products in a chemical reaction. It uses the coefficients in the balanced chemical equation to form conversion factors between moles of different substances.

• For each 3 moles of ethanol \( \mathrm{C}_2\mathrm{H}_6\mathrm{O} \), 2 moles of dichromate \( \mathrm{Cr}_2\mathrm{O}_7^{2-} \) are required.
• For instance, given 0.0326 moles of ethanol, stoichiometry shows you need \( \frac{2}{3} \times 0.0326 = 0.0217 \) moles of dichromate.
• This calculation helps determine the reactant requirement for predefined amounts of products, integral for problem-solving.

Breathalyzer Test Chemistry

The Breathalyzer test relies on a chemical reaction where ethanol is oxidized. The test involves measuring alcohol concentration in the breath and indirectly in blood.

• Ethanol in breath reacts with potassium dichromate in acidic conditions, reducing chromium from +6 to +3 oxidation state.
• The color change from orange to green due to \( \mathrm{Cr}^{3+} \) ions helps measure alcohol levels.
• Understanding stoichiometry ensures the test's accuracy by calculating the amount of dichromate required to oxidize the detected ethanol.

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions involve the transfer of electrons between substances. Identifying oxidation and reduction helps understand chemical changes in reactions.

• In the Breathalyzer test, ethanol \( \mathrm{C}_2\mathrm{H}_6\mathrm{O} \) is oxidized to acetic acid \( \mathrm{C}_2\mathrm{H}_4\mathrm{O}_2 \).
• Concurrently, chromium ions in \( \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \) are reduced from an oxidation state of +6 to +3.
• These complementary processes are crucial for calculating reaction outcomes, such as the amount of acetic acid produced.